# Video: Finding the Derivative of a Polar Equation

Given a polar curve defined by 𝑟 = 𝑓(θ), form an expression for the slope of the curve d𝑦/d𝑥 in terms of 𝜃 and 𝑓.

03:37

### Video Transcript

Given the polar curve defined by 𝑟 is equal to 𝑓 of 𝜃, form an expression for the slope of the curve the derivative of 𝑦 with respect to 𝑥 in terms of 𝜃 and 𝑓.

The question gives us a polar curve defined by the equation 𝑟 is equal to 𝑓 of 𝜃. And it’s asking us to form an expression for the slope of the curve. That’s the derivative of 𝑦 with respect to 𝑥. And it wants this in terms of 𝜃 and 𝑓. To do this, we recall the chain rule, which tells us that the derivative of 𝑦 with respect to 𝑥 is equal to the derivative of 𝑦 with respect to 𝜃 multiplied by the derivative of 𝜃 with respect to 𝑥. So, this is the first step. This gives us an equation for our slope function d𝑦 by d𝑥.

We now recall our polar equations, 𝑦 is equal to 𝑟 sin 𝜃 and 𝑥 is equal to 𝑟 cos 𝜃. We recall the question is asking us to write our slope function in terms of only 𝜃 and 𝑓. So, it would be a good idea to rewrite any equations in terms of 𝜃 and 𝑓 where possible. Luckily, we can use the substitution 𝑟 is equal to 𝑓 of 𝜃 to rewrite these polar equations.

At this point, we could calculate the derivative of 𝑦 with respect to 𝜃 without any problems. However, we can’t directly calculate the derivative of 𝜃 with respect to 𝑥, since we’re given 𝑥 as a function of 𝜃. We can get around this by recalling that multiplying by the derivative of 𝜃 with respect to 𝑥 is the same as dividing by the derivative of 𝑥 with respect to 𝜃. So, we can find an expression for our slope function d𝑦 by d𝑥 by first calculating d𝑦 by d𝜃 and then dividing this by d𝑥 by d𝜃. So, to calculate the derivative of 𝑦 with respect to 𝜃, we differentiate 𝑓 of 𝜃 multiplied by sin 𝜃 with respect to 𝜃.

We can differentiate this by recalling the product rule, which tells us the derivative of the product of two functions 𝑢 and 𝑣 is equal to 𝑢 prime multiplied by 𝑣 plus 𝑣 prime multiplied by 𝑢. So, we first differentiate our function 𝑓 of 𝜃 to get 𝑓 prime of 𝜃 and multiply this by sin 𝜃. Then, we differentiate sin 𝜃, which we recall is cos 𝜃. And we multiply this by 𝑓 of 𝜃. We can then do the same to calculate the derivative of 𝑥 with respect to 𝜃. So, that’s the derivative of 𝑓 of 𝜃 multiplied by the cos of 𝜃 with respect to 𝜃.

We differentiate 𝑓 of 𝜃 to get 𝑓 prime of 𝜃 and then multiply this by the cos of 𝜃. And we add onto this the derivative of the cos of 𝜃, which we recall is negative sin 𝜃. And we multiply this by 𝑓 of 𝜃. We’re now ready to substitute this into our equation for our slope function. We have that d𝑦 by d𝑥 is equal to d𝑦 by d𝜃, divided by d𝑥 by d𝜃. Substituting the equation we got for the derivative of 𝑦 with respect to 𝜃 gives us the numerator of 𝑓 prime of 𝜃 multiplied by sin 𝜃 plus 𝑓 of 𝜃 multiplied by the cos of 𝜃.

And then, substituting the equation we got for the derivative of 𝑥 with respect to 𝜃 gives us a denominator of 𝑓 prime of 𝜃 multiplied by the cos of 𝜃 minus 𝑓 of 𝜃 multiplied by the sin of 𝜃. And so, we have found the formula for our slope function d𝑦 d𝑥 in terms of the function 𝑓 and 𝜃. And it’s fine to just quote this. We don’t need to derive this every time. And so, in conclusion, we have shown that if we are given a polar curve defined by the equation 𝑓 of 𝜃. Then, we can rewrite the slope function the derivative of 𝑦 with respect to 𝑥 as 𝑓 prime of 𝜃 multiplied by sin 𝜃 plus 𝑓 of 𝜃 multiplied by the cos of 𝜃 all divided by 𝑓 prime of 𝜃 multiplied by the cos of 𝜃 minus 𝑓 of 𝜃 multiplied by the sin of 𝜃.